Question

What is the arclength of `f(x)=sqrt((x+3)(x/2-1))+5x` on `x in [6,7]`?

Answer 1

`L~~1.33`

Explanation:

The arclength formula is

`L=int_a^b sqrt(1+f'(x)) dx`

First, find the derivative of the `f(x)`
`d/dx(sqrt((x+3)(x/2-1))+5x)`
`d/dx(sqrt((x+3)(x/2-1)))+d/dx(5x)`
`d/dx(sqrt((x+3)(x/2-1)))+5`

For the first term, let `u=(x+3)(x/2-1)=1/2 x^2+1/2 x -3`
`(du)/dx =x+1/2`

Using `u`-substitution, find `d/dx(sqrt((x+3)(x/2-1)))`

`d/dx(sqrt(u))=1/(2sqrt(u)) (du)/dx`

Replacing these values with `x`, gives
`d/dx(sqrt(u))=(x+1/2)/(2sqrt(1/2 x^2+1/2 x -3))`

Plug this result into the arc-length formula above
`L=int_a^b sqrt(1+f'(x)) dx`
`L=int_6^7 sqrt(1+(x+1/2)/(2sqrt(1/2 x^2+1/2 x -3))) dx`

This is a very difficult function to integrate, so using numeric methods is recommended. Using Simpson's Rule (with `n=10` and `Deltax=1//10`), you could calculate

`L~~(1//10)/3[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+cdots`
`cdots+4f(x_10)+f(x_11)]~~1.33`